TS EAMCET · Maths · Three Dimensional Geometry
If the cartesian equation of the plane passing through the point \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and parallel to the vectors \(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\) is \(a x+b y+c z=1\), then \(18(a+b+c)\) is equal to
- A -3
- B 3
- C 4
- D -4
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
Equation of any plane passing through \((1,2,1)\) is \( \begin{gathered} A(x-1)+B(y-2)+C(z-1)=0 \\ \Rightarrow A x+B y+C z=A+2 B+C \\ \Rightarrow\left(\frac{A}{A+2 B+C}\right) x+\left(\frac{B}{A+2 B+C}\right) y \\ +\left(\frac{C}{A+2 B+C}\right) z=1 \end{gathered} \) According…
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